Is $R$ an integral domain?

Question

Let $R = \{a + b\alpha |\ a,b \in \mathbb{Z}\}\subseteq \mathbb{C}$ where $\alpha = \frac{1}{2}(1+\sqrt{-19})$

To show whether or not $R$ is an integral domain, letting $x = \{a+b\alpha\}$ and $y = \{c+d\alpha\}$, where $a,b,c,d \in \mathbb{Z}$ and showing $xy = yx \in R$ would suffice without going into further, since an integral domain is a commutative ring at first?

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Hint $\ $ One easily verifies that it is a subring of $\,\Bbb C\,$ since it satisfies the subring test, i.e. it contains $\,1\,$ and is closed under subtraction and multiplication (use $\,\alpha^2 = m\alpha + n\,$ for some $\,m,n\in\Bbb Z).\,$ Therefore, being a subring of a field, it is an integral domain.

edited Apr 13 '17 at 12:20

Community

1

answered Feb 15 '15 at 17:30

Bill Dubuque

272,048

That's not quite a hint, more of a direct answer! Good call though, that's a bit easier than the elementary number theory involved in my answer. – walkar Feb 15 '15 at 17:31
@Bill Dubquqe Thanks for your hint/direct answer. – Jellyfish Feb 15 '15 at 17:36
@walker, this problem is just some random problem that I am working on to prepare for my test, not some hw problem :) – Jellyfish Feb 15 '15 at 17:37
Can we just tell it contains 1 just because $1 = 1 + 0*\alpha$??? Kind of arbitrary don't you think? – Jellyfish Feb 15 '15 at 17:41
1

@abcd1234 That's exactly why it contains $1$! Not arbitrary at all, it's an honest-to-goodness element of $R$. – walkar Feb 15 '15 at 17:45
1

@abcd1234 $ $ As you note, $,1\in R,$ follows immediately from the definition of $,R\ \ $ – Bill Dubuque Feb 15 '15 at 17:48
:) thanks both of you for all your help. – Jellyfish Feb 15 '15 at 17:48
@abcd1234 The same idea works for $,\alpha,$ any algebraic integer, i.e. a root of a monic $,f(x)\in\Bbb Z[x].\ \ $ – Bill Dubuque Feb 15 '15 at 17:50

score 1 · Answer 2 · answered Feb 15 '15 at 16:56

1

Hint: An integral domain is a commutative ring with no zero divisors. Write out an arbitrary product of elements in $R$ and set it equal to zero, and see what you can do from there.

It does not suffice to show simply that $R$ is a commutative ring -- that is necessary but not sufficient to being an integral domain.

answered Feb 15 '15 at 16:56

walkar

3,844

That's where I was confused. – Jellyfish Feb 15 '15 at 17:00
A subring of a commutative ring is commutative, so you get that part for free anyway. – Max Feb 15 '15 at 17:01
A subring of an integral domain (such as the field $\mathbf C$) is an integraldomain. – Bernard Feb 15 '15 at 17:07
@abcd1234 Write $(a+b\alpha)(c+d\alpha) = ac+bd\alpha^2 +(ad+bc)\alpha = 0$ and consider all the cases. You will end up with one of $a+b\alpha$ or $c+d\alpha$ must be $0$, or some contradictions with $\alpha \in \mathbb{Z}$. – walkar Feb 15 '15 at 17:07
So, let me get this clear... to be an integral domain, it is always the case that if $xy = 0$ then either $x = 0$ or $y=0$, and we have to prove that true, right? – Jellyfish Feb 15 '15 at 17:17
@abcd1234 Yes. In this instance, to prove $R$ is an integral domain you must prove an arbitrary vanishing product of elements implies one of them must have been 0. – walkar Feb 15 '15 at 17:18
1

Ok, so if $\alpha^2 = \alpha - 5$ then, $ac + bd(\alpha - 5) +(ad +bc)\alpha = (ac - 5bd) + (bd + ad + bc)\alpha =$ some $p + q\alpha$, since $a,b,c,d \in \mathbb{Z}$. Now, what are we concluding here? – Jellyfish Feb 15 '15 at 17:23
@abcd1234 Set $(ac-5bd) + (bd+ad+bc)\alpha = 0$. Then we must have $ac-5bd = 0$ and $(bd+ad+bc)=0$ (as $ac-5bd \neq -(bd+ad+bc)\alpha$ as that implies $\alpha \in \mathbb{Z}$). So $ac = 5bd$ and $bd = -ad-bc$... See how the logic goes? You'll end up with several cases, and some will be contradictions and others will tell you that $a+b\alpha = 0$ or $c+d\alpha = 0$. Good call on recognizing $\alpha^2 = \alpha -5$, I just worked with $\alpha^2$ arbitrarily. – walkar Feb 15 '15 at 17:29
Thanks, I think I got it now :) – Jellyfish Feb 15 '15 at 17:31

Is $R$ an integral domain?

2 Answers2

Linked